Answer
$\frac{11}{18}$
Work Step by Step
$\frac{1}{n(n+3)}$ = $\frac{A}{n}+\frac{B}{n+3}$
$1$ = $A(n+3)+Bn$
$A$ = $\frac{1}{3}$
$B$ = $-\frac{1}{3}$
$\frac{1}{n(n+3)}$ = $\frac{1}{3}(\frac{1}{n}-\frac{1}{n+3})$
$\Sigma_{n=1}^{\infty}\frac{1}{n(n+3)}$ = $\Sigma_{n=1}^{\infty}\frac{1}{3}(\frac{1}{n}-\frac{1}{n+3})$
$S_{N}$ = $\frac{1}{3}(1-\frac{1}{4})+\frac{1}{3}(\frac{1}{2}-\frac{1}{5})+\frac{1}{3}(\frac{1}{3}-\frac{1}{6})+...+\frac{1}{3}(\frac{1}{N}-\frac{1}{N+3})$ = $\frac{11}{18}-\frac{1}{3}(\frac{1}{N+1}+\frac{1}{N+2}+\frac{1}{N+3})$
$\lim\limits_{N \to \infty}S_{N}$ = $\lim\limits_{N \to \infty}[\frac{11}{18}-\frac{1}{3}(\frac{1}{N+1}+\frac{1}{N+2}+\frac{1}{N+3})]$ = $\frac{11}{18}$
